A247698 Brady numbers: B(n) = B(n - 1) + B(n - 2) with B(1) = 2308 and B(2) = 4261.

2308, 4261, 6569, 10830, 17399, 28229, 45628, 73857, 119485, 193342, 312827, 506169, 818996, 1325165, 2144161, 3469326, 5613487, 9082813, 14696300, 23779113, 38475413, 62254526, 100729939, 162984465, 263714404, 426698869, 690413273, 1117112142, 1807525415, 2924637557, 4732162972, 7656800529

Offset: 1

Author

Sebastian Zimmer, Sep 22 2014

Keywords

• nonn
• easy

Comments

B(n) / B(n - 1) approaches the golden ratio as n approaches infinity.

Links

• Logan Cooper, Table of n, a(n) for n = 1..1000 (truncated from 9966 to 1000 terms by _M. F. Hasler_, May 10 2017)
• Brady Haran and Matt Parker, Brady Numbers, Numberphile video (2014).
• Index entries for linear recurrences with constant coefficients, signature (1,1).

Programs

Haskell

brady = let makeSeq a b = a : makeSeq b (a + b) in makeSeq 2308 4261

Magma

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(2308+1953*x)/(1-x-x^2))); // G. C. Greubel, Sep 07 2018

Maple

Brady1 := proc(n::posint)
option remember, system;
if n = 1 then
  2308
elif n = 2 then
  4261
else
  thisproc( n - 1 ) + thisproc( n - 2 )
end if
end proc:
seq( Brady1( n ), n = 1 .. 100 );
# James McCarron, Oct 05 2019
# alternate program
Brady2 := ( n :: posint ) -> coeff( series(x*(2308+1953*x)/(1-x-x^2),x,n+1), x^n ):
seq( Brady2( n ), n = 1 .. 100 );
# James McCarron, Oct 05 2019

Mathematica

LinearRecurrence[{1, 1}, {2308, 4261}, n]
Rest[CoefficientList[Series[x*(2308+1953*x)/(1-x-x^2), {x,0,50}], x]] (* G. C. Greubel, Sep 07 2018 *)

PARI

Vec(-x*(1953*x+2308)/(x^2+x-1) + O(x^50)) \\ Colin Barker, Sep 23 2014

PARI

a(n)=([1,1;1,0]^n*[1953;355])[1,1] \\ Charles R Greathouse IV, Jan 20 2016

Python

def A247698_list(n):
    list = [2308, 4261] + [0] * (n - 2)
    for i in range(2, n):
        list[i] = list[i - 1] + list[i - 2]
    return list
print(A247698_list(32)) # M. Eren Kesim, Jun 28 2021

Formula

a(n) = a(n-1) + a(n-2).

G.f.: x*(2308 + 1953*x) / (1-x-x^2). - Colin Barker, Sep 23 2014

a(n) = k*phi^n + o(1), where k = 976.5 + sqrt(354578.45) = 1571.96.... - Charles R Greathouse IV, Sep 28 2014

a(n) = 2308*A000045(n-2) + 4261*A000045(n-1) = 1953*A000045(n+1) + 355*A000045(n). - M. F. Hasler, May 10 2017

a(n) = F(n+17) - F(n+8) - 9*F(n) - F(n-14) for F(n) = A000045(n). - Greg Dresden, Jul 07 2022